Chapter 6: Chemical Kinetics
Exhaustive Theory, Rate Laws, Derivations, Numericals & Solved PYQs
1. Introduction and Rate of Reaction
Chemical Kinetics is the branch of physical chemistry that deals with the study of rates of chemical reactions, the factors affecting them, and the mechanism by which the reactions proceed.
Average and Instantaneous Rate
- Average Rate: The change in concentration of a reactant or product divided by the time interval over which the change occurs. It is measured over a finite time interval ($\Delta t$).
- Instantaneous Rate: The rate of a reaction at a specific instant of time. It is given by the derivative of concentration with respect to time ($dc/dt$).
Expression for Rate of Reaction
For a general reaction: $aA + bB \rightarrow cC + dD$
The instantaneous rate of reaction is given by:
$$\text{Rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = +\frac{1}{c} \frac{d[C]}{dt} = +\frac{1}{d} \frac{d[D]}{dt}$$
Negative sign indicates the decrease in concentration of reactants, and positive sign indicates the increase in concentration of products. The unit of rate is $\text{mol L}^{-1} \text{s}^{-1}$.
2. Rate Law and Rate Constant
Rate Law: An experimentally determined mathematical expression which relates the rate of a chemical reaction to the molar concentrations of the reactants.
For a reaction $aA + bB \rightarrow \text{Products}$, the rate law is:
$$\text{Rate} \propto [A]^x [B]^y$$ $$\text{Rate} = k [A]^x [B]^y$$Where:
- $k$ is the Rate Constant or Velocity Constant or Specific Reaction Rate.
- $x$ and $y$ are the orders of reaction with respect to A and B. They are determined experimentally and may or may not be equal to the stoichiometric coefficients $a$ and $b$.
Rate Constant ($k$): If $[A] = 1 \text{ mol L}^{-1}$ and $[B] = 1 \text{ mol L}^{-1}$, then $\text{Rate} = k$. Thus, the rate constant is the rate of reaction when the concentration of each reactant is unity.
3. Order and Molecularity of Reaction
Order of Reaction
The sum of the powers of the concentration terms in the experimentally determined rate law expression is called the overall order of the reaction. (Overall Order $n = x + y$).
Order can be an integer (0, 1, 2, 3) or a fraction.
Molecularity of Reaction
The number of reactant molecules, atoms, or ions taking part in an elementary (single-step) reaction, which must collide simultaneously to bring about a chemical reaction.
Distinction between Order and Molecularity
| Order of Reaction | Molecularity of Reaction |
|---|---|
| It is an experimentally determined quantity. | It is a theoretical concept. |
| It can be zero, fraction, or integer. | It is always a whole number (1, 2, 3...) and never zero or a fraction. |
| It applies to elementary as well as complex reactions. | It applies only to elementary reactions. Complex reactions have no overall molecularity. |
| Derived from the rate law expression. | Derived from the balanced equation of an elementary step. |
4. Integrated Rate Laws
Differential rate laws are integrated to obtain equations that give the relationship between the concentration of reactants and time.
A. Integrated Rate Law for First-Order Reaction
For a reaction $A \rightarrow \text{Products}$, Rate = $-\frac{d[A]}{dt} = k[A]^1$
Rearranging: $\frac{d[A]}{[A]} = -k \, dt$
Integrating between limits: at $t=0, [A]=[A]_0$ and at $t=t, [A]=[A]_t$.
$$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt$$
$$\ln[A]_t - \ln[A]_0 = -kt \implies \ln \frac{[A]_t}{[A]_0} = -kt$$
Converting natural log to base-10 log ($\ln x = 2.303 \log_{10} x$):
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
Units of $k$ for first order: $\text{time}^{-1}$ (e.g., $\text{s}^{-1}, \text{min}^{-1}$).
B. Integrated Rate Law for Zero-Order Reaction
For a reaction $A \rightarrow \text{Products}$, Rate = $-\frac{d[A]}{dt} = k[A]^0 = k$
Rearranging: $d[A] = -k \, dt$
Integrating gives: $[A]_t - [A]_0 = -kt$
$$k = \frac{[A]_0 - [A]_t}{t}$$
Units of $k$ for zero order: $\text{mol L}^{-1} \text{time}^{-1}$.
5. Half-Life Period ($t_{1/2}$) of a Reaction
The time required for the initial concentration of the reactant to reduce to exactly half of its original value is called the half-life period of the reaction.
At $t = t_{1/2}$, $[A]_t = \frac{[A]_0}{2}$.
Half-Life of First-Order Reaction
$k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0/2}$
$k = \frac{2.303}{t_{1/2}} \log_{10} 2$
Since $\log_{10} 2 = 0.3010$,
$$t_{1/2} = \frac{0.693}{k}$$
Conclusion: $t_{1/2}$ for a first-order reaction is independent of initial concentration $[A]_0$.
Half-Life of Zero-Order Reaction
$k = \frac{[A]_0 - [A]_0/2}{t_{1/2}}$
$k = \frac{[A]_0 / 2}{t_{1/2}}$
$$t_{1/2} = \frac{[A]_0}{2k}$$
Conclusion: $t_{1/2}$ for a zero-order reaction is directly proportional to initial concentration $[A]_0$.
6. Pseudo-First-Order Reaction
Reactions that are not truly first-order but behave as first-order reactions under certain conditions are called pseudo-first-order reactions.
This occurs when one of the reactants is present in large excess, so its concentration remains practically constant throughout the reaction.
Example: Hydrolysis of ester (Methyl acetate) in acidic medium
$CH_3COOCH_3(aq) + H_2O(l) \xrightarrow{H^+} CH_3COOH(aq) + CH_3OH(aq)$
Expected Rate Law: $\text{Rate} = k'[CH_3COOCH_3][H_2O]$
Since $H_2O$ is the solvent and is present in huge excess, $[H_2O]$ is almost constant. Let $k'[H_2O] = k$.
Observed Rate Law: $\text{Rate} = k[CH_3COOCH_3]$
The true order is 2, but the observed order is 1.
7. Arrhenius Equation and Activation Energy
The rate of a chemical reaction generally increases with temperature. Arrhenius proposed a quantitative relationship between the rate constant ($k$) and temperature ($T$):
$$k = A e^{-E_a / RT}$$
- $A$ = Arrhenius constant (or frequency factor or pre-exponential factor).
- $E_a$ = Activation Energy ($\text{J mol}^{-1}$).
- $R$ = Universal Gas Constant ($8.314 \text{ J K}^{-1}\text{mol}^{-1}$).
- $T$ = Absolute temperature (in Kelvin).
Logarithmic Form for Calculation
Taking natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$
Converting to base-10 log: $\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}$
Comparing $k$ at two different temperatures ($T_1$ and $T_2$):
$$\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$
Effect of Catalyst
A positive catalyst increases the rate of reaction by providing an alternative pathway or mechanism with a lower activation energy ($E_a$). Since $E_a$ is lowered, more reactant molecules cross the energy barrier at the same temperature, increasing the rate constant $k$.
8. Solved Textbook Numericals
Problem 1: First-Order Reaction (Time calculation)
A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period.
Solution:
Let initial concentration $[A]_0 = 100$
Since 30% is decomposed, the amount remaining $[A]_t = 100 - 30 = 70$
Time $t = 40 \text{ min}$
Step 1: Calculate $k$
$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$
$k = \frac{2.303}{40} \log_{10} \frac{100}{70} = \frac{2.303}{40} \log_{10}(1.428)$
$k = \frac{2.303}{40} \times 0.1548 = 0.00891 \text{ min}^{-1}$
Step 2: Calculate $t_{1/2}$
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00891} = 77.78 \text{ min}$
Answer: The half-life period is 77.78 minutes.
Problem 2: Arrhenius Equation
The rate constant of a reaction is $1.2 \times 10^{-3} \text{ s}^{-1}$ at 30°C and $2.1 \times 10^{-3} \text{ s}^{-1}$ at 40°C. Calculate the energy of activation of the reaction. ($R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$)
Solution:
$T_1 = 30^\circ\text{C} = 303 \text{ K}, \quad k_1 = 1.2 \times 10^{-3} \text{ s}^{-1}$
$T_2 = 40^\circ\text{C} = 313 \text{ K}, \quad k_2 = 2.1 \times 10^{-3} \text{ s}^{-1}$
Formula: $\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
$\log_{10} \left( \frac{2.1 \times 10^{-3}}{1.2 \times 10^{-3}} \right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313 - 303}{303 \times 313} \right]$
$\log_{10} (1.75) = \frac{E_a}{19.147} \left[ \frac{10}{94839} \right]$
$0.2430 = \frac{E_a \times 10}{19.147 \times 94839} \implies 0.2430 = \frac{10 \times E_a}{1815882.3}$
$E_a = \frac{0.2430 \times 1815882.3}{10} = 44125.9 \text{ J/mol} = 44.13 \text{ kJ/mol}$
Answer: The activation energy is 44.13 kJ/mol.
9. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. Define: Order of reaction. (March 2013, Oct 2015, March 2022)
Answer: The sum of the powers of the concentration terms of reactants in the experimentally determined rate law expression is called the overall order of the reaction.
Q2. Define: Half-life period of a reaction. (March 2017, March 2020)
Answer: The time required for the initial concentration of the reactant to reduce to exactly half of its original value is called the half-life period ($t_{1/2}$) of the reaction.
Q3. Give the unit of the rate constant for a zero-order reaction. (Oct 2018)
Answer: $\text{mol L}^{-1} \text{time}^{-1}$ (or $\text{mol dm}^{-3} \text{s}^{-1}$).
2 Mark Questions (SA-I)
Q4. Distinguish between order of reaction and molecularity of reaction. (March 2014, Oct 2017, March 2019)
- Order: It is an experimentally determined value. It can be zero, fraction, or a whole number. It is applicable to elementary and complex reactions.
- Molecularity: It is a theoretical concept. It is always a whole number and can never be zero or a fraction. It is applicable only to elementary steps.
Q5. Explain pseudo-first-order reaction with a suitable example. (March 2015, March 2021)
A reaction which has a higher molecularity/true order but follows first-order kinetics under certain conditions (usually when one reactant is in large excess) is called a pseudo-first-order reaction.
Example: Hydrolysis of methyl acetate.
$CH_3COOCH_3 + H_2O \rightarrow CH_3COOH + CH_3OH$
Rate = $k'[CH_3COOCH_3][H_2O]$. Since water is in excess, its concentration is constant. The rate becomes dependent only on the ester: Rate = $k[CH_3COOCH_3]$. Thus, it acts as a first-order reaction.
3 Mark Questions (SA-II)
Q6. Derive the integrated rate law equation for a first-order reaction. (Oct 2013, March 2016, Oct 2020)
1. Consider a first-order reaction: $A \rightarrow \text{Products}$.
2. The differential rate law is: Rate = $-\frac{d[A]}{dt} = k[A]$
3. Rearranging the terms: $\frac{d[A]}{[A]} = -k \, dt$
4. Integrating both sides between limits (at $t=0, [A]=[A]_0$ and at $t=t, [A]=[A]_t$):
$$\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt$$
$$\ln[A]_t - \ln[A]_0 = -k(t - 0)$$
$$\ln \frac{[A]_t}{[A]_0} = -kt \implies kt = \ln \frac{[A]_0}{[A]_t}$$
5. Converting natural logarithm ($\ln$) to base-10 logarithm ($\log_{10}$), we multiply by 2.303:
$$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$$
This is the required integrated rate law for a first-order reaction.
Q7. Show that for a first-order reaction, the time required for 99% completion is twice the time required for 90% completion. (March 2018, March 2023)
For a first-order reaction: $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$
Let $[A]_0 = 100$
Case 1: 99% completion
Amount remaining $[A]_t = 100 - 99 = 1$
$t_{99\%} = \frac{2.303}{k} \log_{10} \frac{100}{1} = \frac{2.303}{k} \log_{10}(10^2) = \frac{2.303}{k} \times 2$ --- (Eq. 1)
Case 2: 90% completion
Amount remaining $[A]_t = 100 - 90 = 10$
$t_{90\%} = \frac{2.303}{k} \log_{10} \frac{100}{10} = \frac{2.303}{k} \log_{10}(10^1) = \frac{2.303}{k} \times 1$ --- (Eq. 2)
Conclusion
Comparing Eq. 1 and Eq. 2:
$t_{99\%} = 2 \times \left( \frac{2.303}{k} \times 1 \right) = 2 \times t_{90\%}$
Hence proved: $t_{99\%} = 2 \times t_{90\%}$.
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